Cosets and Normal Subgroups

Cosets and Normal Subgroups (Last Updated: November 3, 2017) These notes are derived primarily from Abstract Algebra, Theory and Applications by Thomas Judson (16ed). Most of this material is drawn from Chapters 6, and 9-11. 1. Cosets Cosets are arguably one of the strangest structures that students encounter in abstract algebra, along with factor groups, which are strongly related. Here s a motivating question for this section: if H is a subgroup of a group G, then how are H and G related? A partial answer to this is contained in Lagrange s Theorem. Definition. Let G be a group and H a subgroup of G. A left coset of H with representative in G is the set gh = {gh : h H}. A right coset of H with representative in G is the set Hg = {hg : h H}. Disclaimer/Warning: For abelian groups there is no difference between left and right cosets. In additive notation, we write g + H = {g + h : h H}. Note that additive groups are by definition abelian and so g + H = H + g. Disclaimer/Warning: Cosets are NOT subgroups in general! Example. Let H = 3 = {0, 3} in Z 6. The cosets are 0 + H = {0, 3} = 3 + H 1 + H = {1, 4} = 4 + H 2 + H = {2, 5} = 5 + H. Example. Let K = {(1), (1 2)} in S 3. The left cosets are (1)K = (1 2)K = {(1), (1 2)} (1 3)K = (1 2 3)K = {(1 3), (1 2 3)} (2 3)K = (1 3 2)K = {(2 3), (1 3 2)}

The right cosets are K(1) = K(1 2) = {(1), (1 2)} K(1 3) = K(1 3 2) = {(1 3), (1 3 2)} K(2 3) = K(1 2 3) = {(2 3), (1 2 3)} Note that, except for the coset of the elements in H, the left and right cosets are different. Example. Let L = {(1), (1 2 3), (1 3 2)} in S 3. The left cosets are The right cosets are (1)L = (1 2 3)L = (1 3 2)L = L (1 2)L = (1 3)L = (2 3)L = {(1 2), (1 3), (2 3)} L(1) = L(1 2 3) = L(1 3 2) L(1 2) = L(1 3) = L(2 3) = {(1 2), (1 3), (2 3)} In this case, the left and right cosets are the same. Lemma 1. Let H be a subgroup of G and suppose g 1, g 2 G. The following are equivalent. (1) g 1 H g 2 H (2) g 1 H = g 2 H (3) Hg1 1 = Hg2 1 (4) g 2 g 1 H (5) g 1 1 g 2 H. Proof. (1) (2) Assume g 1 H g 2 H. We claim the opposite inclusion holds. Let g 2 h g 2 H for some h H. Since g 1 g 1 H g 2 H, then g 1 = g 2 h for some h. But then g 2 = g 1 (h ) 1 and so g 2 h = g 1 (h ) 1 h = g 1 ((h ) 1 h) g 1 H. (2) (3) Assume g 1 H = g 2 H. Let hg1 1 Hg1 1 with h H. Note that g 1 g 1 H = g 2 H, so g 1 = g 2 h for some h H. Then g1 1 = (h ) 1 g2 1 and so hg 1 1 = h((h ) 1 g 1 2 ) = (h(h ) 1 )g 1 2 Hg 1 2. Thus, Hg1 1 Hg2 1. A similar proof shows the reverse inclusion and the result follows. (3) (4) Assume Hg1 1 = Hg2 1. Then g 1 2 Hg2 1 = Hg1 1 so g2 1 = hg1 1 for some h H. Thus, g 2 = g 1 h 1 g 1 H. (4) (5) Assume g 2 g 1 H. Then g 2 = g 1 h for some h H. Thus, g 1 1 g 2 = h H. (5) (1) Assume g1 1 g 2 H. Let g 1 h g 1 H for some h H. By our assumption, g1 1 g 2 = h for some h H. Thus, g 1 = g 2 (h ) 1 and g 1 h = g 2 ((h ) 1 h) g 2 H. Therefore g 1 H g 2 H.

Disclaimer/Warning: Remember when we proved that the equivalence classes under some equivalence relation partition the set. This is when we need that result. Theorem 2. Let H be a subgroup of a group G. The left cosets of H in G partition G. Proof. Define an equivalence relation on G by g 1 g 2 if (and only if) g 2 g 1 H. Clearly is reflexive because g 1 = g 1 e g 1 H, so g 1 g 1. Suppose g 1 g 2. Then g 2 g 1 H. By the previous lemma, this implies g 1 H = g 2 H and so g 1 = g 1 e g 1 H = g 2 H. Thus, g 2 g 1 and so is reflexive. Finally, suppose g 1 g 2 and g 2 g 3. Then g 2 g 1 H and g 3 g 2 H. Again using the above lemma, g 1 H = g 2 H and g 2 H = g 3 H, so g 1 H = g 3 H. Thus, g 3 g 1 H so g 1 g 3. The equivalence classes under this relation are the cosets and therefore a partition of G. Note that the above result holds if we replace left with right. One could prove it similarly or apply the lemma (exercise). Definition. Let G be a group and H a subgroup. The index of H in G is the number of left cosets in G, denoted [H : G]. We will show momentarily that the number of left cosets is equal to the number of right cosets. Example. In the previous examples, we have [Z 6 : H] = 3, [S 3 : K] = 3, and [S 3 : L] = 2. Theorem 3. Let H be a subgroup of G. The number of left cosets of H in G is the same as the number of right cosets. Proof. Denote by L H the set of left cosets of H in G and by R H the set of right cosets of H in G. We will establish a bijection between these sets, which shows that L H = R H. Define a map φ : L H R H gh Hg 1. First we need to show that this map is well-defined. Suppose g 1 H = g 2 H. By the lemma, φ(g 1 H) = Hg 1 1 = Hg 1 2 = φ(g 2 H). Thus, φ is well-defined. Suppose φ(g 1 H) = φ(g 2 H), then Hg 1 1 = Hg 1 2. Again, the lemma implies g 1H = g 2 H, so φ is injective. Let Hg R H. Then φ(g 1 H) = H(g 1 ) 1 = Hg, so φ is surjective. Thus, φ is bijective and so the result holds. 2. Lagrange s Theorem The next lemma proves a fact we observed in the examples.

Lemma 4. Let H be a subgroup of G. For all g G, H = gh. Proof. Fix g G and define a map φ : H gh h gh It is clear that φ is well-defined. We will show that it is bijective. Suppose φ(h) = φ(h ). Then gh = gh and so by left cancelation, h = h. Thus φ is injective. Next let gh gh, then φ(h) = gh so the map is surjective. Thus, φ is a bijection and so H = gh. The proof of Lagrange s Theorem is very simple but only because we ve done a sizable amount of legwork already. Theorem 5 (Lagrange s Theorem). Let G be a finite group and H a subgroup of G. G / H = [G : H]. In particular, the order of H divides the order of G. Then Proof. The group G is partitioned into [G : H] distinct left cosets. Each coset has exactly H elements by the previous lemma. Hence, G = H [G : H]. We ll now examine a host of consequence of Lagrange s Theorem. Corollary 6. Suppose G is a finite group and g G. Then (1) g G. (2) g G = e. Proof. (1) Note that g = g. Thus, we simply apply Lagrange s Theorem with H = g. (2) Set d = G. Then g = k d by part (1). Thus, d = kl for some integer l. Then g d = g kl = (g k ) l = e l = e. Corollary 7. Let G be a group with G = p, p prime. Then G is cyclic and any g G, g e, is a generator. Proof. Let g G, g e. Then g G by the previous corollary, so g = 1 or p. But g e so g > 1, so g = p. Thus, g = G. Let s take a moment to consider what we have just proved. The last corollary says that every group of prime power order is cyclic. Thus, if G is a group of order p, p prime, then G is essentially the same as Z p. We will formalize this in coming sections. Corollary 8. Let H, K be subgroups of a finite group G such that K H G. Then [G : K] = [G : H][H : K].

Proof. We have by Lagrange s Theorem, [G : K] = G K = G H H = [G : H][H : K]. K The converse of Lagrange s Theorem is false (in general). If n G, this does not imply that there exists a subgroup H of G with H = n. For example, A 4 = 12 but A 4 (the alternating group on 4 letters) has no subgroup of order 6. Now for a fun little diversion into number theory. Definition. The Euler φ-function is defined as φ : N N with φ(1) = 1 and φ(n) = {m N : 1 m < n and gcd(m, n) = 1}. This next theorem formalizes something we discussed much earlier. Theorem 9. Let n N. (1) U(n) = φ(n). (2) (Euler s Theorem) Let a, n be integers such that n > 0 and gcd(a, n) = 1. Then a φ(n) 1 mod n. (3) (Fermat s Little Theorem) Let p be any prime and suppose p a. Then a p 1 1 mod p. Furthermore, for any b Z, b p b mod p. Proof. (1) By definition, U(n) is the set of integers m, 1 m < n that are invertible (under multiplication) mod n. We need only show that any such number is relatively prime to n. Suppose mk 1 mod n for some integer k. Then mk = nl + 1, or mk + n( l) = 1 for some integer l. Thus, by the Euclidean Algorithm, gcd(m, n) = 1. (2) Since U(n) = φ(n), then a φ(n) = 1 for all a U(n). This is equivalent to a φ(n) 1 mod n. (3) Apply Euler s Theorem with n = p. 3. Normal subgroups and factor groups Disclaimer/Warning: We are very much deviating from the structure of the book at this point. I make no apologies. On one hand, normality encapsulates the idea of left and right cosets being the same, but more importantly they lead to the idea of factor groups. The motivation for this is simply that subgroups do not give a full sense of the structure of the groups. The factor groups give the complementary hidden structure. Definition. A subgroup N of a group G is normal if gn = Ng for all g G.

Example. (1) If G is abelian, then every subgroup is normal. (2) K = {(1), (1 2)} in S 3 is not normal. (3) L = {(1), (1 2 3), (1 3 2)} is normal. Theorem 10. Let G be a group and N a subgroup. The following are equivalent. (1) N is normal. (2) For all g G, gng 1 N. (3) For all g G, gng 1 = N. Proof. (1) (2) Assume N is normal and fix g G. Let n N, then by the definition of normal, gn = Ng so gn = n g for some n N. Thus, gng 1 = (n g)g 1 = n (gg 1 ) = n N, so gng 1 N. (2) (3) Assume gng 1 N for all g G. We claim the opposite inclusion holds (for all g G). Let n N and fix g G. By our assumption, g 1 ng N (since it holds for all g G). Thus, g 1 ng = n for some n N, so n = gn g 1 gng 1. Therefore, N gng 1. (3) (1) Assume gng 1 = N for all g G. We will show gn Ng. The proof that Ng gn is similar. Let g G and n N. Then gng 1 = n for some n N. Thus, gn = n g Ng. Let N be a normal subgroup of a group G. We define a binary operation on the cosets by (an)(bn) = (ab)n. Before we proceed, we should verify that this operation is actually well-defined. Said another way, we must verify that the product is independent of choice of coset representative. Let an = cn and bn = dn. We claim (an)(bn) = (cn)(dn). Since c an and d bn, then there exists n 1, n 2 N such that c = an 1 and d = bn 2. Thus (cd)n = (an 1 )(bn 2 )N = (an 1 )b(n 2 N) = (an 1 )(bn) = (an 1 )(Nb) = a(n 1 N)b = anb = (ab)n. Note that bn = Nb by normality. Disclaimer/Warning: The operation for additive cosets is equivalent, (a + N) + (b + N) = (a + b) + N. We denote by G/N the cosets of N in G. Theorem 11. Let N be a normal subgroup of G. The cosets of N in G form a group G/N (with the operation above) of order [G : N]. Proof. We have already shown that the operation is binary. We verify the group axioms. Throughout, let an, bn, cn G/N. The operation is associative since (an)[(bn)(cn)] = (an)(bcn) = (a(bc))n = ((ab)c)n = (abn)(cn) = [(an)(bn)](cn).

The identity element in G/N is just the coset of the identity in G. (en)(an) = (ea)n = an. Finally, the inverse of an is just the coset of a 1, (an)(a 1 N) = (aa 1 )N = en. Disclaimer/Warning: We typically denote the identity element en simply by N. Definition. Let G be a group and N a normal subgroup. The group G/N is the factor group of G by N. Example. Let G = S 3 and N = {(1), (1 2 3), (1 3 2)}. (Note that N = A 3 ). Then N is normal in G (we have already verified this) and the cosets are N and (1 2)N. The Cayley Table is given by N (1 2)N N N (1 2)N (1 2)N (1 2)N N Observe that this table is equivalent to that of Z 2. Example. Consider the subgroup H = 3Z in Z. There are 3 cosets: 0 + H, 1 + H, 2 + H. Write out the Cayley Table for Z/H and observe how it is equivalent to that of Z 3. Example. Consider D n generated by r, s with r n = id, s 2 = id, srs = r 1. Let R n be the subgroup of rotational symmetries. Then R n is normal in D n because srs = r 1 R n. Note that D n /R n = [D n : R n ] = D n / R n = 2n/n = 2. Thus, the Cayley Table is equivalent to Z 2. 4. Homomorphisms Our next goal will be to explain how normal subgroups and factor groups arise naturally in the study of groups. In particular, a normal subgroups can be thought of as the kernel, or left over, part of a map between two groups. Definition. A homomorphism is a function φ : (G, ) (H, ) between groups such that φ(g 1 g 2 ) = φ(g 1 ) φ(g 2 ) for all g 1, g 2 G. The set im φ = {φ(g) : g G} is called the image of φ.

One should think of a homomorphism as a structure preserving map, that is, it preserves the structure of the groups G and H. In fact, whenever you see the word morphism, this is generally what is meant (though not always for groups). Example. (1) φ : GL 2 (R) R given by φ(m) = det(m) is a homomorphism. Note that in this case φ is surjective but not injective. (2) φ : M 2 (R) R given by φ(m) = det(m) is not a homomorphism because det(m + N) det(m) + det(n) in general. (3) φ : Z 2 Z 4 by φ(0) = 0 and φ(1) = 2. This map is injective but not surjective. (4) Choose g G, G a group, and define φ : Z G by φ(n) = g n. Then φ is a homomorphism and im φ = g. (5) Let C = ({1, 1}, ). Define φ : S n C by σ(σ) = sgn σ where sgn σ = 1 if σ is even and 1 is σ is odd. (We say sgn σ is the sign of σ). (6) Let G be a group and N a normal subgroup. The map φ : G G/N given by φ(g) = gn. This map is surjective but not injective. (This map is known as the quotient map. Proposition 12. Let φ : G H be a homomorphism of groups. (1) φ(e G ) = e H. (2) For any g G, φ(g 1 ) = g 1. (3) If K is a subgroup of G, then φ(k) is a subgroup of H. (4) If L is a subgroup of H, then φ 1 (L) = {g G : φ(g) L} is a subgroup of G. Furthermore, if L is normal in H then φ 1 (L) is normal in G. Proof. (1) Let g G, then φ(g)e H = φ(g) = φ(ge G ) = φ(g)φ(e G ). By left cancellation, e H = φ(e G ). (2) Let g G, then by (1), e H = φ(gg 1 ) = φ(g)φ(g 1 ). By uniqueness of the inverse, φ(g 1 ) = φ(g) 1. (3) Let K be a subgroup of G. Since e G K, the K. Let a, b φ(k), so there exists k 1, k 2 K such that φ(k 1 ) = a and φ(k 2 ) = b. Then ab 1 = φ(k 1 )φ(k 2 ) 1 K because K is a subgroup of G. Thus, φ(k) is a subgroup of H. (4) Let L be a subgroup of H. Since e G φ 1 (L), then φ 1 (L). Choose c, d φ 1 (L), so φ(c), φ(d) L. Then by (2) φ(cd 1 ) = φ(c)φ(d 1 ) = φ(c)φ(d) 1 L because L is a subgroup of H. Thus, cd 1 φ 1 (L) and so φ 1 (L) is a subgroup of G.

Finally, suppose L is normal in H. Let g H. We claim gφ 1 (L)g 1 φ 1 (L). Let a φ 1 (L), then φ(gag 1 ) = φ(g)φ(a)φ(g) 1 L because L is normal. Thus, gag 1 φ 1 (L) for all a φ 1 (L), so φ 1 (L) is normal in G. Definition. The kernel of a homomorphism φ : G H is the set ker φ = {g G : φ(g) = e}. Theorem 13. Let φ : G H be a group homomorphism. Then ker φ is a normal subgroup of G. Proof. By the previous proposition, e ker φ and so ker φ. Let g 1, g 2 ker φ. Then Thus, ker φ is a subgroup. φ(g 1 g 1 2 ) = φ(g 1)φ(g 2 ) 1 = ee 1 = e. Next we claim ker φ is normal. Let g G and k ker φ. Then φ(gkg 1 ) = φ(g)φ(k)φ(g) 1 = φ(g)eφ(g) 1 = e. Thus, g ker φg 1 ker φ and so ker φ is normal. Example. (1) φ : GL 2 (R) R given by φ(m) = det(m). Then ker φ = SL 2 (R). (2) φ : Z 2 Z 4 by φ(0) = 0 and φ(1) = 2. We can check directly that ker φ = {0, 2}. (3) Choose g G, G a group, and define φ : Z G by φ(n) = g n. If g = then ker φ = {0}. If g = k, then ker φ = kz. (4) Let φ : S n C by σ(σ) = sgn σ. Then ker φ = A n. (5) Let G be a group and N a normal subgroup. The map φ : G G/N given by φ(g) = gn has ker φ = N. 5. Isomorphisms Definition. Two groups (G, ) and (H, ) are said to be isomorphic if there exists a bijective homomorphism φ : G H. The map φ in this case is called an isomorphism. Example. (1) Let H be a cyclic group of order 3 generated by h (so H = {e, h, h 2 }). The map φ : Z 3 H is an isomorphism. Note that we have already verified that this map is an isomorphism, we need only verify that it is injective and surjective, but this is an easy exercise in this case. (2) The groups D 3 and S 3 are isomorphic. This is not difficult to see by comparing the Cayley Tables of the two groups. Theorem 14. Let φ : G H be an isomorphism of groups. (1) φ 1 : H G is an isomorphism.

(2) G = H. (3) If G abelian, then H is abelian. (4) If G is cyclic, then H is cyclic. (5) If G has a subgroup of order n, then H has a subgroup of order n. Proof. (1) By standard results on functions, φ 1 exists and is bijective (see the book s introductory chapter). We need only verify that it is a homomorphism, but this is easy. Since φ(ab) = φ(a)φ(b) for all a, b G, then φ 1 (ab) = φ 1 (a)φ 1 (b) for all a, b G. (2) This follows because φ is bijective. (3) Assume G is abelian. Let h 1, h 2 H, then there exist (unique) elements g 1, g 2 G such that φ(g 1 ) = h 1 and φ(g 2 ) = h 2. Then h 1 h 2 = φ(g 1 )φ(g 2 ) = φ(g 1 g 2 ) = φ(g 2 g 1 ) = φ(g 2 )φ(g 1 ) = h 2 h 1. Thus, H is abelian. (4) Exercise. (5) Let K be a subgroup of G of order n. Then φ(k) is a subgroup of H because φ is a homomorphism. Moreover, φ(k) = n because φ is bijective. Proposition 15. A surjective group homomorphism φ : G H is an isomorphism if and only if ker φ = {e H }. Proof. Assume φ is an isomorphism. Because it is a homomorphism, e H ker φ. Because it is 1-1, then ker φ = {e H }. Assume ker φ = {e H }. We claim φ is 1-1. Suppose φ(g 1 ) = φ(g 2 ) for some g 1, g 2 G. Then e = φ(g 1 )φ(g 2 ) 1 = φ(g 1 g 1 2 ). Thus, by our assumption, g 1g 1 2 = e so g 1 = g 2. It follows that G is 1-1 and thus an isomorphism. The next theorem may be regarded as a classification of all cyclic groups (up to isomorphism). Theorem 16. Let G be a cyclic group with generator a G. (1) If a =, then G = Z. (2) If a = n <, then G = Z n. Proof. Define a map φ : Z G by φ(n) = a n. Note that we have already verified that φ is a homomorphism. We verify in (1) that φ is bijective and leave (2) as an exercise. It is clear that φ is surjective. We will show that it is injective and thus an isomorphism.. Let k, l Z such that φ(k) = φ(l). Then a k = a l and so a k l = e. Since a =, this implies k l = 0, so k = l. Thus, φ is injective and the result follows.

Corollary 17. If G = p, p prime, then G = Z p. Proof. Let a G, a e. Then by Lagrange s Theorem, a p. Since a 1 then a = p. Thus, G = a. By the previous theorem, G = Z p. Let G be a group and for each g G define a map λ g : G G a ga. Exercise. Show that λ g is an isomorphism and hence a permutation of G. Theorem 18 (Cayley s Theorem). Every group is isomorphic to a group of permutations. Proof. Exercise. See below. Exercise. Let G be a group and G = {λ g : g G}. Define a map Φ : G G g λ g. Show that Φ is an isomorphism. 6. Direct products You have already been introduced to one type of product group, the external direct product. Recall that if (G, ) and (H, ) are groups, then G H = {(g, h) : g G, h H} is a group under the operation below, (g 1, h 1 ) (g 2, h 2 ) = (g 1 g 2, h 1 h 2 ) for all g 1, g 2 G, h 1, h 2 H. Recall that if G = p, p prime, then G = Z p. Here is a related result. Proposition 19. Let m, n be positive integers, then Z m Z n = Zmn if and only if gcd(m, n) = 1. Proof. Exercise. Definition. If G is a group and A, B subgroups of G such that G = A B, then G is said to be the external direct product of A and B. Example. The cyclic group Z 6 has subgroups A = 2 and B = 3. Note that 2 = Z 2 and 3 = Z 2. By the above proposition, Z 6 = Z2 Z 3. Thus, Z 6 is the external direct product of A and B.

Now we introduce a new type of subgroup product. For a group G with subgroups H, N, we define the set HN = {hn : h H, n N}. Note that in general this is not a subgroup of G. Certainly it is if G is abelian. The next proposition gives another criteria. Proposition 20. Let G be a group with subgroups H, N. If N is normal, then HN is a subgroup of G. Proof. Clearly e HN because e H N. Let h 1 n 1, h 2 n 2 HN. Then (h 1 n 1 )(h 2 n 2 ) 1 = (h 1 n 1 )(n 1 2 h 2). Because N is normal, Nh 2 = h 2 N. Thus, n 1 2 h = hn 3 and n 1 h = hn 4 for some n 3, n 4 N. We therefore have (h 1 n 1 )(h 2 n 2 ) 1 = (h 1 n 1 )(n 1 2 h 2) = (h 1 n 1 )(h 2 n 3 ) = h 1 (h 2 n 4 )n 3 = (h 1 h 2 )(n 4 n 3 ) HN. Thus, HN is a subgroup of G. Disclaimer/Warning: In additive notation, HN is H + N = {h + n : h H, n N}. Recall that subgroups of abelian groups are always normal and so H + N is always a subgroup of G. Exercise. Let H, N be subgroups of a group G with N normal. Prove that N is normal subgroup of HN and H N is a normal subgroup of H. Definition. A group G is the interal direct product of subgroups H and K provided (1) G = HK (as sets), (2) H K = {e}, and (3) hk = kh for all h H, k K. Example. Let G = U(8), H = {1, 3}, K = {1, 5}. Then G = HK (because 3 5 7 mod 8). Clearly H K = {1}, and because G is abelian we have hk = kh for all h H, k K. Thus, G is the internal direct product of H and K. Example. S 3 is not an internal direct product of its subgroups. Since S 3 = 6 then without loss of generality we need subgroups H, K with H = 3 and K = 2. But then H = Z 3 and K = Z 2. The condition hk = kh for all h H, k K now implies that S 3 is abelian, a contradiction. Lemma 21. If G is the internal direct product of subgroups H and K, then every element in G can be written uniquely as hk for some h H, k K. Proof. Let g G. By the first condition of internal direct products, G = HK, and so g = hk for some h H, k K. We claim this decomposition is unique. Suppose g = h k for some h H, k K. By the second condition, hk = h k (h ) 1 h = k k 1 H K = {e}.

Thus, (h ) 1 h = e and k k 1 = e, so h = h and k = k. The next theorem says that if G is the internal direct product of subgroups H and K, then it is also the external direct product of those subgroups. Theorem 22. If G is the internal direct product of subgroups H and K, then G = H K. Proof. Define a map φ : H K G (h, k) hk. By the lemma, φ is bijective. We need only show that it is a homomorphism. Let (h 1, k 1 ), (h 2, k 2 ) H K. Then by the third condition of internal direct products, φ((h 1, k 1 )(h 2, k 2 )) = φ(h 1 h 2, k 1 k 2 ) = (h 1 h 2 )(k 1 k 2 ) = (h 1 k 1 )(h 2 k 2 ) = φ(h 1, k 1 )φ(h 2, k 2 ). Thus, φ is a homomorphism and therefore an isomorphism. Example. By a previous example, U(8) = Z 2 Z 2. Exercise. Prove that D 6 = S3 Z 2. 7. The isomorphism theorems The following theorems explain how factor group structures fit into the overall picture of group structure. They are also incredibly powerful tools for proving that two groups are isomorphic. Our next result actually generalizes an earlier result that a homomorphism is an isomorphism if and only if its kernel is trivial. The first isomorphism theorem says that the factor group of a group by the kernel of an homomorphism is isomorphic to the image of the homomorphism. Theorem 23 (First Isomorphism Theorem). Let φ : G H be a homomorphism and K = ker φ. There exists an isomorphism ψ : G/K φ(g) gk φ(g). That is, G/K = φ(g). Proof. First recall that by an earlier exercise, K is a normal subgroup and thus G/K is well-defined. However, ψ is a map defined on a set of cosets and therefore we must check that it is well-defined. Suppose g 1 K = g 2 K for some g 1 K, g 2 K G/K. We must show that ψ(g 1 K) = ψ(g 2 K). Because K is normal, our hypothesis implies that g 1 2 g 1 K. Therefore, ψ(g 1 2 g 1) = e, so ψ(g 2 ) = φ(g 1 ). But then ψ(g 1 K) = φ(g 2 K). Thus, ψ is well-defined.

We now check that ψ is a homomorphism. Let g 1 K, g 2 K G/K. Then ψ((g 1 K)(g 2 K)) = ψ((g 1 g 2 )K) = φ(g 1 g 2 ) = φ(g 1 )φ(g 2 ) = ψ(g 1 K)ψ(g 2 K). It is left to show that ψ is bijective. Clearly ψ is surjective since for all φ(g) φ(g), ψ(gk) = φ(g). To show injectivity we reverse the argument above for well-definedness. Suppose ψ(g 1 K) = ψ(g 2 K) for some g 1 K, g 2 K G/K. Then φ(g 1 ) = φ(g 2 ) so g2 1 g 1 K. Because K is normal, g 1 K = g 2 K, so ψ is injective. Example. Consider the map φ : Z Z 3 given by φ(k) = k mod 3. This map is a surjective homomorphism (check!) and ker φ = 3Z. By the first isomorphism theorem, Z/3Z = Z 3. Exercise. Generalize the previous example (and carefully check all steps) to show that Z/nZ = Z n for all positive integers n. Theorem 24 (Second Isomorphism Theorem). Let H, N be subgroups of a group G with N normal. Then H/H N = HN/N. Proof. By an earlier exercise, N is a normal subgroup of HN and H N is a normal subgroup of H. Thus, the given factor groups are well-defined. Define a map φ : H HN/N h hn. We claim this map is a surjective homomorphism. Let h 1, h 2 H. Then by the normality of N, φ(h 1 h 2 ) = (h 1 h 2 )N = h 1 (h 2 N) = h 1 (Nh 2 ) = (h 1 N)(Nh 2 ) = (h 1 N)(h 2 N) = φ(h 1 )φ(h 2 ). Thus, φ is a homomorphism. Let hnn HN/N, then clearly hnn = hn = φ(h), so φ is surjective. By the first isomorphism theorem, H/ ker φ = HN/N, so we need only show that ker φ = H N. Let x H N, then φ(x) = xn = N because x N, so H N ker φ. On the other hand, if y ker φ, then yn = φ(y) = N, so y N. Thus, ker φ H N so the sets are equal. Disclaimer/Warning: In additive notation, the theorem is rephrased as H + N instead of HN. Example. Let H = 2Z and N = 3Z be subgroups of Z. Clearly N is normal because Z is abelian. Then H N = 6Z and H + N = Z. Thus, by the second isomorphism theorem, 2Z/6Z = Z 3. The next two results are also important, but in the interest of time and sanity I will leave the proofs as reading exercises. However, I suspect that the third isomorphism theorem is one that you can prove once you fully understand the proofs above, especially that of the second isomorphism theorem.

Theorem 25 (Correspondence Theorem). Let N be a normal subgroup of a group G. Then there is a bijection from the set of all subgroups containing N and the set of subgroups of G/N. Furthermore, the normal subgroups of G containing N correspond to normal subgroups of G/N. Example. Consider the group G = Z 12 and let N = 4. Note that G/N = Z 4. Then we can identify the subgroups of G containing N with the subgroups of G/N. subgroups of G containing N subgroups of G/N N {N} {0, 2, 4, 6, 8, 10, 12} {N, 2 + N} G {N, 1 + N, 2 + N, 3 + N}. Theorem 26 (Third Isomorphism Theorem). Let H, N be normal subgroups of a group G with N H. Then G/H = (G/N)/(H/N). Example. By the third isomorphism theorem, Z/3Z = (Z/6Z)/(3Z/6Z)